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Chapter 15, Problem 1.
Find the Laplace transform of:
(a) cosh at (b) sinh at
[ Hint: cosh x = ( )
x x e e
−
2
, sinh x = ( )
x x e e
− − 2
.]
Chapter 15, Solution 1.
(a) 2
e e cosh(at)
at - at
=
[ ] =
s a
s a
L cosh(at) 2 2 s a
s
(b) 2
e e sinh(at)
at - at − =
[ ] =
s a
s a
L sinh(at) 2 2 s a
a
Chapter 15, Problem 2.
Determine the Laplace transform of:
(a) cos( ω t + θ) (b) sin( ω t + θ)
Chapter 15, Solution 2.
(a) f (t)=cos(ωt)cos(θ)−sin(ωt)sin(θ)
F( s)= cos(θ)L^ [^ cos(ωt)]^ −sin(θ)L[^ sin(ωt)]
F( s) = 2 2 s
scos( ) sin( )
θ −ω θ
(b) f (t)=sin(ωt)cos(θ)+cos(ωt)sin(θ)
F( s)= sin(θ)L^ [^ cos(ωt)]^ +cos(θ)L[^ sin(ωt)]
F( s) = 2 2 s
ssin() cos( )
θ −ω θ
e tu ( t )
t sin 4
− 2
(c) e tu ( ) t
t cosh 2
− 3 (d) e tu ( t )
t sinh
− 4
(e) te tu ( ) t
t sin 2
−
Chapter 15, Solution 3.
(a) [^ e cos( 3 t)u(t)]^ =
-2t L (s 2 ) 9
s 2 2
(b) [^ e sin( 4 t)u(t)]^ = -2t L (s 2 ) 16
2
(c) Since [^ ]^2
s a
s cosh(at ) −
L =
[ (^) e cosh( 2 t)u(t)] (^) = -3t L (s 3 ) 4
s 3 2
(d) Since [^ ]^
2 2 s a
a sinh(at ) −
L =
[ e sinh(t)u(t)] =
-4t L (s 4 ) 1
2
(e) [^ ]^
(s 1 ) 4
e sin( 2 t) 2
L =
If f (t) ←⎯→ F(s)
F(s ) ds
Thus, [^ ]^ [^ (^ ) ]
- t 2 -^1 2 (s 1 ) 4 ds
- d L t e sin( 2 t) = + +
2 (s 1 ) ((s 1 ) 4 )
2 2
L [^ t e-tsin( 2 t)]^ = 2 2 ((s 1 ) 4 )
4 (s 1 )
t e u ( t )
4 2 t 3
−
(c) ( ) ( ) t
dt
d
2 tu t − 4 δ (d)
( ) e u ( ) t
t 1 2
−−
(e) 5 u ( t 2 ) (f) e u ( t )
t 3 6
−
(g) ( ) t
dt
d n
n
Chapter 15, Solution 5.
(a) [^ ]^
s 4
scos( 30 ) 2 sin( 30 ) cos( 2 t 30 ) 2
L + ° =
[ ]
s 4
scos( 30 ) 1
ds
d t cos( 2 t 30 ) 2 2
2 2 L
2 -^1 s 1 s 4 2
ds
d
ds
d
2 -^12 -^2 s 1 s 4 2
s 4 2 s 2
ds
d
( )
2 3
2
2 2 2 2 2 2 s 4
s 1 2
( 8 s)
s 4
2 s
s 4
s 1 2
s 4
2 3
2
2 2 s 4
s 1 2
( 8 s)
s 4
2 3
3 2
2 3
2
s 4
4 3 s 8 s
s 4
(-3 3 s 2)(s 4 )
[ (^) t cos( 2 t+ 30 °)] (^) = 2 L ( ) 2 3
2 3
s 4
8 12 3 s 6 s 3 s
(b) [ ] =
5
4 - 2t
(s 2 )
L 3 t e 3 5 (s 2 )
(c) = − ⋅ − = ⎥ ⎦
− δ 4 (s 1 0 ) s
(t) dt
d L 2 tu(t) 4 2 4 s s
(d) 2 e u(t) 2 e u(t) -(t- 1) -t =
L [^2 e-(t-1)u(t)]^ =
s 1
2 e
(e) Using the scaling property,
[ ] = ⋅ ⋅ = ⋅ ⋅ =
2 s
s ( 12 )
L 5 u(t 2 ) 5 s
(f) [^ ]^ =
s 13
6 e u(t)
(g) Let f (t)= δ(t). Then, F( s)= 1.
⎥= − − ′ −^ L
δ
− − f(t) s F(s) s f( 0 ) s f( 0 ) dt
d (t) dt
d (^) n n 1 n 2 n
n
n
n
L L
⎥= ⋅ − ⋅ − ⋅ −^ L
δ
− − f(t) s 1 s 0 s 0 dt
d (t) dt
d (^) n n 1 n 2 n
n
n
n
L L
⎥^ =
δ (t) dt
d n
n
L
n s
Chapter 15, Problem 8.
Find the Laplace transform F ( s ), given that f ( t ) is:
(a) 2 tu ( t − 4 )
(b) 5 cos( ) t δ( t − 2 )
(c) e u ( t t )
t −
−
(d) sin( 2 t ) ( ut −τ)
Chapter 15, Solution 8.
(a) 2t=2(t-4) + 8
f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4)
4 4 4 2 2
s s s F s e e e s s s s
− − ⎛^ ⎞ −
(b)
2
0 0
( ) ( ) 5cos ( 2) 5cos 5cos 2 2
st st st s F s f t e dt t t e dt te e t
∞ ∞ − − − − = = − = = =
(c)
t ( t ) e e e
− − − τ −τ
( ) ( ) ( )
t f t e e u t
τ τ
− − − = −
( 1) 1 ( ) 1 1
s s e F s e e s s
τ τ τ
− + − − = =
(d) sin 2 t = sin[2( t − τ ) + 2 ]τ = sin 2( t − τ ) cos 2τ + cos 2( t −τ )sin 2τ
f t ( ) = cos 2 sin 2( τ t − τ ) ( u t −τ ) + sin 2τ cos 2( t − τ ) ( u t −τ)
2 2
( ) cos 2 sin 2 4 4
s s s F s e e s s
τ τ τ τ
− − = +
5cos(2)e
–2s
Chapter 15, Problem 9.
Determine the Laplace transforms of these functions:
(a) f ( ) t = ( t − 4 ) ( ut − 2 )
(b) ( ) 2 ( 1 )
4 = −
− g t e ut
t
(c) h ( ) t = 5 cos( 2 t − 1 ) ( ) ut
(d) p ( ) t = 6 [ u ( t − 2 ) − u ( t − 4 )]
Chapter 15, Solution 9.
(a) f (t)=(t− 4 )u(t− 2 )=(t− 2 )u(t− 2 )− 2 u(t− 2 )
F( s) = 2
-2s
2
-2s
s
2 e
s
e −
(b) g( t) 2 e u(t 1 ) 2 e e u(t 1 )
-4t -4 -4(t-1) = − = −
G (s) = e (s 4 )
2 e
4
-s
(c) h( t)= 5 cos( 2 t− 1 )u(t)
cos(A −B)=cos(A)cos(B)+sin(A)sin(B )
cos( 2 t− 1 )=cos( 2 t)cos( 1 )+sin( 2 t)sin( 1 )
h( t)= 5 cos( 1 )cos( 2 t)u(t)+ 5 sin( 1 )sin( 2 t)u(t )
s 4
5 sin( 1 ) s 4
s H (s) 5 cos( 1 ) 2 2
H( s) = s 4
s 4
2. 702 s
2 2
(d) p( t)= 6 u(t− 2 )− 6 u(t− 4 )
P (s) =
- 2s -4s e s
e s
